I do not understand why do we divide with the k! that is used in nCk when multiplying two combinations or more combinations that has the same k value

Question

I recently started studying Permutations and Combinations and following is one interesting question I came across.

There are 8 students in a class. The class teacher wants to divide divide those students into four teams. The sizes of teams need not all be equal and a team may consist of even one person. Show that the required team can be formed in 1701 ways.

How I approached the problem

We can select students in following 5 ways,

1. 5 Students, 1 Student, 1 Student, 1 Student
2. 4 Students, 2 Students, 1 Student, 1 Student
3. 3 Students, 3 Students, 1 Student, 1 Student
4. 3 Students, 2 Students, 2 Students, 1 Student
5. 2 Students, 2 Students, 2 Students, 2 Students

Then we can find combinations as below for the given five cases,

1. $\binom{8}{5} \cdot \binom{3}{1} \cdot \binom{2}{1} \cdot \binom{1}{1}$
2. $\binom{8}{4} \cdot \binom{4}{2} \cdot \binom{2}{1} \cdot \binom{2}{1}$
3. $\binom{8}{3} \cdot \binom{5}{3} \cdot \binom{2}{1} \cdot \binom{2}{1}$
4. $\binom{8}{3} \cdot \binom{5}{2} \cdot \binom{3}{2} \cdot \binom{1}{1}$
5. $\binom{8}{2} \cdot \binom{8}{2} \cdot \binom{8}{2} \cdot \binom{8}{2}$

But above answer has been wrong and the correct answer is

1. $\binom{8}{5}$
2. $\binom{8}{4} \cdot \binom{4}{2}$
3. $\binom{8}{3} \cdot \binom{5}{3}$ divided by 2!
4. $\binom{8}{3} \cdot \binom{5}{2} \cdot \binom{3}{2}$ divided by 2!
5. $\binom{8}{2} \cdot \binom{8}{2} \cdot \binom{8}{2} \cdot \binom{8}{2}$ divided by 4!

Now I have 2 questions,

1. How come second one is correct? Why aren't we choosing in the scenarios where 1 student is present?
2. Why are we dividing with 2! and 4! etc... in second secnario?

More importantly how to recognise when to divide with 2! and 4! etc in problems...?

Answer 1

You need to be very clear in the concept for such problems.

Basically it depends whether teams are labeled (eg team A, team B, etc) or unlabeled.

A further very important point to note is that even if teams have not been given names, they can become labeled by size eg three teams of 4, 3,1, are automatically labeled because you can distinguish them by their size.

To come to your problem, teams have not been given names (labels) but they can become labeled by size.

Take case $3$ with two teams of $3$ Suppose people $A,C,F$ are in one team and $B,D,X$ in the other team. Now you could interchange them to $B,D,X = ACF$ and it is the same two teams, so we need to divide by $2!$

Similarly, if $3$ teams have the same size, we need to divide by $3!$